A 10g bullet travelling at 200m/s strike and remains embedded in a 2kg target which is originally at rest but free to move .At what speed does the target move off?
Answers
Answer: 0.99m/s
Explanation:
According to the law of conservation of momentum,
For a collision occurring between two bodies or objects, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
(The body is in isolated system)
Therefore
Total momentum before collision = Total momentum after collision
LHS=
Total momentum before collision
= m1u1+m2u2
=(0.01*200)+(2*0)
= 2kgm/s
Total momentum after collision
= m1v1+m2v2
= (0.01+2)*v
According to the law,
2kgm /s = (2.01) v
Therefore v= 2/2.01
= 0.99m/s
The target will move at a speed off 0.99m/s.
Answer:
Explanation:
Answer: 0.99m/s
Explanation:
According to the law of conservation of momentum,
For a collision occurring between two bodies or objects, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
(The body is in isolated system)
Therefore
Total momentum before collision = Total momentum after collision
LHS=
Total momentum before collision
= m1u1+m2u2
=(0.01*200)+(2*0)
= 2kgm/s
Total momentum after collision
= m1v1+m2v2
= (0.01+2)*v
According to the law,
2kgm /s = (2.01) v
Therefore v= 2/2.01
= 0.99m/s
The target will move at a speed off 0.99m/s.